Question

A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into another circular orbit of radius 1.01 R around the earth. The period of revolution of the second satellite is larger than that of the first one by a percent of (approximately)

• A. 0.5
• B. 1.0
• C. 1.5
• D. 3.0

Hint:

For problems involving orbital periods and radii, remember Kepler's Third Law ($T^2 \propto R^3$). When dealing with small percentage changes, the binomial approximation $(1+x)^n \approx 1 + nx$ is a very useful tool for quick calculations.

Answer 1

The Correct Option is C

Step 1: Apply Kepler's Third Law of Planetary Motion.
Kepler's Third Law states that for any satellite orbiting a central body (like the Earth), the square of its orbital period ($T$) is directly proportional to the cube of the semi-major axis (which is the radius $R$ for a circular orbit). Mathematically, this can be written as: \[ T^2 \propto R^3 \] Or, $T = C R^{3/2}$, where $C$ is a constant. Step 2: Set up the ratio of periods for the two satellites.
Let $T_1$ be the period of the first satellite and $R_1$ its orbital radius. Let $T_2$ be the period of the second satellite and $R_2$ its orbital radius. From the problem statement: $R_1 = R$ $R_2 = 1.01 R$ Using Kepler's Third Law for both satellites: $T_1^2 = C R_1^3$ $T_2^2 = C R_2^3$ Now, take the ratio of the squares of the periods: \[ \frac{T_2^2}{T_1^2} = \frac{C R_2^3}{C R_1^3} = \left(\frac{R_2}{R_1}\right)^3 \] Substitute the given values for $R_1$ and $R_2$: \[ \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{1.01 R}{R}\right)^3 \] \[ \left(\frac{T_2}{T_1}\right)^2 = (1.01)^3 \] Step 3: Calculate the ratio of the periods.
Take the square root of both sides: \[ \frac{T_2}{T_1} = (1.01)^{3/2} \] Since $0.01 \ll 1$, we can use the binomial approximation $(1+x)^n \approx 1 + nx$ for small $x$. Here, $x=0.01$ and $n=3/2$. \[ (1.01)^{3/2} = (1 + 0.01)^{3/2} \approx 1 + \left(\frac{3}{2}\right)(0.01) \] \[ \frac{T_2}{T_1} \approx 1 + 1.5 \times 0.01 \] \[ \frac{T_2}{T_1} \approx 1 + 0.015 \] \[ \frac{T_2}{T_1} \approx 1.015 \] Step 4: Calculate the percentage increase in the period.
The percentage increase is given by: \[ \text{Percentage Increase} = \left(\frac{T_2 - T_1}{T_1}\right) \times 100% \] \[ = \left(\frac{T_2}{T_1} - 1\right) \times 100% \] Substitute the calculated ratio: \[ = (1.015 - 1) \times 100% \] \[ = 0.015 \times 100% \] \[ = 1.5% \] The final answer is $\boxed{1.5}$.